Question

An object 2 cm in size is placed 30cm in front of a concave mirror of focal length 15cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? What will be the nature and the size of the image formed? Draw a ray diagram to show the formation of the image in this case.

Answer 1

According to the question; 

Object distance (u) = -30cm; 

Focal length (f) = -15cm; 

Image distance = v; 

By mirror formula;

\(\frac1v+\frac1u=\frac1f\)

⇒ \(\frac1v+\frac{1}{-30}=\frac{1}{-15}\)

⇒ \(\frac1v=\frac{1}{30}-\frac{1}{15}\)

⇒ \(\frac1v=\frac{1-2}{30}\)

\(\frac1v\) = \(-\frac1{30}\)

⇒ v = 30cm.

Thus, screen should be placed 30cm in front of the mirror (Centre of curvature) to obtain the real image. 

Height of object h₁= 2cm;

Magnification = \(\frac{h_2}{h_1}=-\frac{v}u\)

Putting values of v and u

Magnification = \(\frac{h_2}{2}=-\frac{-30}{-30}\)

⇒ \(\frac{h_2}{2}=-1;\)

⇒ h₂ = 2 x -1 = -2.

Height of image is 2 cm. 

Negative sign means image is inverted. 

Thus real, inverted image of size same as that of object is formed. 

Diagram below shows the image formation.